Two branches, and a constant difference.
The foci lie outside the vertices here, the opposite of an ellipse. That is exactly what c² = a² + b² forces: c must exceed a.
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- The Hyperbola
The Hyperbola
The ellipse adds the two distances. The hyperbola subtracts them.
Find the foci
x²16 − y²9 = 1
For a hyperbola c² = a² + b² = 16 + 9 = 25, so c = 5 and the foci are (±5, 0). C subtracts, which is the ellipse's rule. D gives the vertices. The x² term is the positive one, so the branches open left and right and the foci sit on the x-axis.
The POSITIVE term decides which way it opens.
x²36 − y²64 = 1
opens left and right
x² is positive
y²64 − x²36 = 1
opens up and down
y² is positive
a² is under the positive term
always
even if it is the smaller number
Size does not decide anything here
For an ellipse the LARGER denominator gave the major axis. For a hyperbola it is the SIGN that decides, and a² can perfectly well be smaller than b². Carrying the ellipse habit across is the commonest error in this topic.
Everything follows from a, b and c.
a² = 36, b² = 64
a under the positive term
vertices (±6, 0)
a from the centre
c² = 36 + 64 = 100, c = 10
ADD for a hyperbola; foci (±10, 0)
asymptotes y = ±(86)x = ±(43)x
slope ±ba
For a horizontal hyperbola the asymptote slopes are ±ba. For a vertical one they are ±ab. Sketch the rectangle 2a by 2b around the centre and the asymptotes are its diagonals.
The branches approach the asymptotes without meeting them.
near the vertex
curve is well inside
rounded turn
far out
almost straight
hugging the asymptote
ever touching?
no
same idea as a rational graph
Drawing the asymptotes first, lightly, makes a hyperbola easy to sketch. Without them the branches tend to come out looking like parabolas, which they are not.
Variations: vertical slopes, not equal to 1, and backwards.
Each one uses the same a, b and c, with one detail turned round.
Asymptotes of a vertical hyperbola: y²25 − x²4 = 1
a² = 25 under the positive y², b² = 4
a = 5, b = 2
slope = ±ab for a vertical one
rise is along the y-axis now
y = ±(52)x
the 2a by 2b box is 4 wide and 10 tall
For a vertical hyperbola the slope flips to ab. Sketching the box settles it: rise over run.
Not equal to 1 yet: 4x² − 9y² = 36
x²9 − y²4 = 1
divide every term by 36
c² = 9 + 4 = 13
add
foci (±√13, 0)
an irrational c is fine
Divide by the constant first; the 4 and 9 in front are not a² and b².
Backwards: vertices (±3, 0), foci (±5, 0)
a = 3, c = 5, horizontal
both pairs on the x-axis
b² = c² − a² = 25 − 9 = 16
rearrange c² = a² + b²
x²9 − y²16 = 1
minus sign on the y term
Going backwards we subtract to find b², but the curve is still a hyperbola, so the equation has a minus.
A shifted centre, and a vertical opening.
centre (1, −3)
signs flip
y² positive, so it opens up and down
a² = 4 even though 5 is bigger
c² = 4 + 5 = 9, c = 3
add
foci (1, −3 ± 3), so (1, 0) and (1, −6)
along the vertical axis
Note that a² = 4 is smaller than b² = 5, and nothing is wrong. The positive term owns a², full stop. An ellipse could never be written that way round, which is one quick way to tell the two apart.
On the exam
Tests give an equation and ask which way it opens, its foci, or its asymptotes, often placing it next to an ellipse with the same numbers. The fastest route: the positive term names the axis and owns a², add for c², and read the slopes from the 2a by 2b box. A 10-second check: c must come out larger than a, so the foci sit outside the vertices.
Now you try
Practice problem
Find the foci
y²9 − x²16 = 1
The y² term is positive, so the branches open up and down and the foci sit on the y-axis. c² = 9 + 16 = 25 gives c = 5, so the foci are (0, ±5). A puts them on the x-axis because 16 is the larger number — but for a hyperbola the SIGN decides, not the size. D subtracts, which is the ellipse rule.
Hint: ignore which denominator is bigger. Find the term with the plus sign in front and everything follows from that.